Given an array of n elements such that every element of array is an integer in the range 1 to n, find the sum of all the distinct elements of the array.

Examples:

Input : arr[] = {5, 1, 2, 4, 6, 7, 3, 6, 7} Output : 28 The distinct elements in the array are 1, 2, 3, 4, 5, 6, 7 Input: arr[] = {1, 1, 1} Output: 1

The problem has appeared here as a general problem and the solution will work for the above case also. But a better approach is explained below.

The approach is to mark the occurrences of the array elements by making the elements at those indices as negative. Example, a[0] = 1, a[1] = 1, a[2] = 1.

We check if a[abs(a[i])-1] is >=0, if yes, mark a[abs(a[i])-1] as negative. i.e. a[0] = 1 >=0, we mark a[1-1] as a[0] = -1. Next, a[1], check if (abs(a[1]-1) is +ve or not. If -ve, it means a[1] has already occurred before, else it is the first occurrence of this element. Refer the below code.

## C++

`// C++ program to find sum of distinct elements` `#include <iostream>` `using` `namespace` `std;` ` ` `// Returns sum of distinct elements in arr[] assuming` `// that elements in a[] are in range from 1 to n.` `int` `sumOfDistinct(` `int` `a[], ` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// If element appears first time` ` ` `if` `(a[` `abs` `(a[i]) - 1] >= 0) {` ` ` `sum += ` `abs` `(a[i]);` ` ` `a[` `abs` `(a[i]) - 1] *= -1;` ` ` `}` ` ` `}` ` ` ` ` `return` `sum;` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `int` `a[] = { 5, 1, 2, 4, 6, 7, 3, 6, 7 };` ` ` `int` `n = ` `sizeof` `(a)/` `sizeof` `(a[0]);` ` ` `cout << sumOfDistinct(a, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// JAVA program to find sum of distinct ` `// elements in sorted order` `import` `java.io.*;` `import` `java.util.*;` `import` `java.math.*;` ` ` `class` `GFG{` ` ` ` ` `// Returns sum of distinct elements in arr[]` ` ` `// assuming that elements in a[] are in ` ` ` `// range from 1 to n.` ` ` `static` `int` `sumOfDistinct(` `int` `a[], ` `int` `n)` ` ` `{` ` ` `int` `sum = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` ` ` `// If element appears first time` ` ` `if` `(a[Math.abs(a[i]) - ` `1` `] >= ` `0` `) {` ` ` `sum += Math.abs(a[i]);` ` ` `a[Math.abs(a[i]) - ` `1` `] *= -` `1` `;` ` ` `}` ` ` `}` ` ` ` ` `return` `sum;` ` ` `}` ` ` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `a[] = { ` `5` `, ` `1` `, ` `2` `, ` `4` `, ` `6` `, ` `7` `, ` `3` `, ` `6` `, ` `7` `};` ` ` `int` `n = a.length;` ` ` `System.out.println(sumOfDistinct(a, n) );` ` ` `}` `}` ` ` `// This code is contributed by Nikita Tiwari.` |

## Python

`# Python program to find sum of distinct elements ` `# in sorted order` `import` `math` ` ` `# Returns sum of distinct elements in arr[]` `# assuming that elements in a[] are in ` `# range from 1 to n.` `def` `sumOfDistinct(a , n) :` ` ` `sum` `=` `0` ` ` `i ` `=` `0` ` ` `while` `i < n:` ` ` ` ` `# If element appears first time` ` ` `if` `(a[` `abs` `(a[i]) ` `-` `1` `] >` `=` `0` `) :` ` ` `sum` `=` `sum` `+` `abs` `(a[i])` ` ` `a[` `abs` `(a[i]) ` `-` `1` `] ` `=` `a[` `abs` `(a[i]) ` `-` `1` `] ` `*` `(` `-` `1` `)` ` ` `i ` `=` `i ` `+` `1` ` ` ` ` `return` `sum` `;` ` ` ` ` `# Driver code` `a ` `=` `[ ` `5` `, ` `1` `, ` `2` `, ` `4` `, ` `6` `, ` `7` `, ` `3` `, ` `6` `, ` `7` `]` `n ` `=` `len` `(a)` `print` `sumOfDistinct(a, n)` ` ` `# This code is contributed by Nikita Tiwari.` |

## C#

`// C# program to find sum of distinct ` `// elements in sorted order` `using` `System;` ` ` `class` `GFG{` ` ` ` ` `// Returns sum of distinct elements` ` ` `// in arr[] assuming that elements` ` ` `// in a[] are in range from 1 to n ` ` ` `static` `int` `sumOfDistinct(` `int` `[]a, ` `int` `n)` ` ` `{` ` ` `int` `sum = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// If element appears first time` ` ` `if` `(a[Math.Abs(a[i]) - 1] >= 0) {` ` ` `sum += Math.Abs(a[i]);` ` ` `a[Math.Abs(a[i]) - 1] *= - 1;` ` ` `}` ` ` `}` ` ` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]a = {5, 1, 2, 4, 6, 7, 3, 6, 7};` ` ` `int` `n = a.Length;` ` ` `Console.Write(sumOfDistinct(a, n));` ` ` `}` `}` ` ` `// This code is contributed by Nitin Mittal` |

## PHP

`<?php` `// PHP program to find sum of` `// distinct elements` ` ` `// Returns sum of distinct ` `// elements in arr[] assuming` `// that elements in a[] are ` `// in range from 1 to n.` `function` `sumOfDistinct(` `$a` `, ` `$n` `)` `{` ` ` `$sum` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{` ` ` ` ` `// If element appears first time` ` ` `if` `(` `$a` `[` `abs` `(` `$a` `[` `$i` `]) - 1] >= 0)` ` ` `{` ` ` `$sum` `+= ` `abs` `(` `$a` `[` `$i` `]);` ` ` `$a` `[` `abs` `(` `$a` `[` `$i` `]) - 1] *= -1;` ` ` `}` ` ` `}` ` ` ` ` `return` `$sum` `;` `}` ` ` ` ` `// Driver code` ` ` `$a` `= ` `array` `(5, 1, 2, 4, 6, 7, 3, 6, 7);` ` ` `$n` `= sizeof(` `$a` `);` ` ` `echo` `sumOfDistinct(` `$a` `, ` `$n` `) ;` ` ` `// This code is contributed by nitin mittal` `?>` |

Output:

28

Time complexity: O(n)

Auxiliary Space : O(1)

This article is contributed by **Ekta Goel**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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